= Exterior quarter-disc hodograph for a downward right-angle turn
{title2=$e^{\pi w}=((q^2-1)/(q^2+1))^2$}
A unit-speed free layer turning downward around a reentrant wall corner has <complex velocity> $q=u-iv$ in the first-quadrant exterior of the <unit disc>. Upstream $q=1$, downstream $q=i$, and the corner has infinite speed because its fluid angle is $3\pi/2$. Squaring $q$, applying $(s-1)/(s+1)$, and squaring again maps this domain to the <upper half-plane> with upstream, corner and downstream at $0,1,\infty$. The potential strip map $e^{\pi w}$ has the same normalization, proving the formula. The sign of the turn and the fluid corner angle determine whether the relevant hodograph is an exterior or interior quarter disc.
Back to article page
